Question 1208825
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I'll focus on problem #1 only.
Problem #2 seems to have a typo in it. I have a feeling it's supposed to say (1+i)^3, but I'm not sure.


Anyway, with problem #1, let's consider i^23 and worry about the negative exponent later.
To evaluate i^23, we divide the exponent by 4 and look at the remainder.
Refer to this <a href="https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1207482.html">similar question</a> to see why we divide by 4.


23/4 = 5 <font color=blue>remainder 3</font>
This means, i^23 = i^<font color=blue>3</font> = -i


Another way we can arrive at that is to say:
i^(23) = i^(20+3)
= i^(20)*i^3
= i^(5*4)*i^3
= (i^4)^5*i^3
= (1)^5*i^3
= i^3
= -i
Notice I rewrote 23 as 20+3. The 20 is the largest multiple of 4 just short of 23.
The useful exponent rules are a^b*a^c = a^(b+c) and (a^b)^c = a^(b*c).


There are probably other ways of determining that i^23 = i^3 = -i.


Now to account for the negative exponent, we could have these steps
i^(-23) = 1/(i^23)
= 1/(-i)
= i/(-i^2)
= i/(-(-1))
= <font color=red>i</font>
The jump from the 2nd line to the 3rd line is when we multiply top/bottom by i. 
Doing this step will convert the imaginary denominator to a real number.



Answer: i^(-23) = <font color=red size=4>i</font>
Verification with <a href="https://www.wolframalpha.com/input?i=i%5E%28-23%29">WolframAlpha</a>
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